# Time, what is time?

So earlier this evening Commander Chris Hadfield, a Canadian astronaut on the ISS, tweeted the question

I gave a quick reply which ended up generating quite a bit of discussion, so I thought I might write up a slightly more detailed answer to the question.

The problem as posed relates to the special theory of relativity. Objects in relative motion will measure time running at a different rate. In this case, the astronauts travelling on the ISS are travelling quite quickly relative to an observer on the ground.

We can measure this time dilation using the Lorentz transformation. We begin by taking the speed of the space station, and use it to calculate a quantity called $\gamma$.

$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}.$

We use the values
$v = 8km/s = 8000m/s\\ c = 300,000,000m/s$

and get

$\gamma = \frac{1}{\sqrt{1 - \frac{8000^2}{300000000^2}}}\\ = \frac{1}{\sqrt{0.9999999992888889}}\\ = 1.0000000003555556\\$

So this is our time dilation factor, which tells us how much slower the clocks on the space station are running.

Now, if the astronauts have been up there for what we on Earth would measure as 5 months, how much time have the astronauts experienced with their slow running clocks? Well, we can work that out with the equation

$t = \frac{t_0}{\gamma}$.

Here we let $t_0$ be 5 months, or $5 \times 30 \times 24 \times 60 \times 60 = 12960000s$. So the time experienced by the astronauts will be

$t = \frac{12960000s}{1.0000000003555556}\\ = 12959999.995392s.$

This is very close to exactly 5 months, but it’s just a tiny bit smaller. How much smaller is it? Well,

$\Delta t_{SR} = t_0 - t\\ = 12960000s - 12959999.995392s\\ = 0.004608s$

This is much less than a single second! In fact, it’s a touch over 4.6 milliseconds.

So there we go, the astronauts will be 4.6 milliseconds younger than they would be if they’d stayed planted on Earth. The time difference is due to time dilation, a direct consequence of their movement relative to Earth and the theory of special relativity.

But this is only part of the picture! My answer to the original question only took into account the relative speeds of the space station and an observer on Earth. If we delve deeper, we find that gravity also has an effect on the rate at which clocks tick. When standing at sea level, we’re around 6371km from the centre of the Earth. But the space station orbits at around 370km above sea level, which puts them 6741km from the centre of the Earth. Because of this difference in distance, the space station experiences a weaker gravity field than we do here on terra-firma.

The effect of gravity on time can be calculated by using the equations from general relativity. Gravity causes time to run more slowly, so we need to work out how strong this effect is both on Earth, and on the space station. We’ll use the following equation to work out the time dilation factor due to gravity.

$\gamma = \frac{1}{\sqrt{1 - \frac{r_0}{R}}}.$

In this equation $r_0$ is the Schwarzschild radius of the Earth, which is based on its gravity. It essentially tells us how far away the event horizon would be if the Earth collapsed into a black hole. It turns out to be quite small, just 9mm, or 0.009 metres.

The variable $R$ is the distance from the centre of the Earth. Based on the numbers above we can work out that for someone standing at sea level we get

$\gamma_E = \frac{1}{\sqrt{1 - \frac{0.009}{6371000}}}\\ = \frac{1}{\sqrt{0.9999999985873489}}\\ = 1.0000000007063257.$

For a person on the space station, we get

$\gamma_S = \frac{1}{\sqrt{1 - \frac{0.009}{6741000}}}\\ = \frac{1}{\sqrt{0.9999999986648865}}\\ = 1.000000000667557.$

We can now work out how long 5 Earth months would take on the space station, by considering the relative clock speeds on Earth and the space station.

$t = t_0\times\frac{\gamma_E}{\gamma_S}\\ = 12960000s \times \frac{1.0000000007063257}{1.000000000667557}\\ = 12960000.000502443s.$

So we see that the effect of general relativity causes the astronauts to experience more time than if they’d stayed on Earth! How much more time?

$\Delta t_{GR} = t_0 - t\\ = 12960000s - 12960000.000502443s\\ = -0.000502443s,$

or around 0.5 milliseconds.

So if we put together the effects of both general relatively (gravity) and special relativity (speed differences) we get the total time difference,

$\Delta t_{tot} = \Delta t_{SR} + \Delta t_{GR}\\ = 0.004608 - 0.000502443\\ = 0.004105557s.$

So there we have it, the answer to the Commander’s question is 4.1 milliseconds! 5 months in space to travel into the future by less than the blink of an eye. Worth it? Totally!

## 5 thoughts on “Time, what is time?”

1. Pingback: Time Dilation | joewalshe.net

2. Thank you for this Tim.

I read in a layman physic book this and it blew my mind whenever I remember :
If you are in a spacecraft accelerating at 1g away from the earth for 10 years, then decelerating at 1g for 10 years, then accelerating at 1g toward the earth for 10 years and finally decelerating at 1g for 10 years.
You end up near the earth 40 years older but the earth is around 59 000 years older.

I was never able to make the computation myself because of the acceleration not equal to 0.

3. Thank you for that very comprehensible explanation! However, I have a problem with it: the assumption that people on earth have v=0 and thus the relative speed of the ISS is 8km/s. I put it to you that the relative speed of the ISS to us on the ground is more complex than that due to more speeds to consider – the rotation of earth around its own axis; the traveling of earth around the sun; the sun traveling through space etc… Now the most tangible problem with the relative speed of ISS vs earth is to me that earth travels around the sun. Similar to the fact that any given point of a wheel on a bicycle moving forward has different speeds during one revolution even if the speed is constant relative to the bicycle. I’m sure you know what I’m talking about. So lets say the ISS was in a equatorial orbit (I know it isn’t but for the same of my point…) and that the earth is going around the sun at a given speed vE. Then the relative speed of the ISS is always changing between vE+8km/s and vE-8km/s? My point is that there is actually no special relativity between earth and ISS. What do you make of that?

Regards from Sweden (pretty much exactly other side of the earth)

• You’re correct in your observation that I have simplified the problem significantly. The rotation around the sun is actually the one thing we don’t really need to worry about here. The ISS and Earth revolve around the Sun at the same speed (vE), and so we can cancel this speed out of all of our equations.

The tricky part is the fact that the ISS is rotating around the Earth, which is in turn spinning on its axis. So for an observer standing on the Earth, the ISS actually appears to move in a somewhat crazy pattern. If we wanted to be completely accurate we’d need to take into account the details of this crazy pattern. As it turns out though, the simplified equations give us an answer which is “close enough”. I left these details out of my post in the interest of simplicity. The fascinating (and frustrating) thing about physics is that for simplified things, the calculations can be quite simple, but for realistic problems, the calculations are often insanely complex.